In both cases, the particle is the circular motion anti-clockwise around the circle. \(t_g\) is perpendicular to the plane of the circle and is in the direction given by the right hand screw rule. \(t_c\) sweeps outwards on the left and \(t_c\) sweeps inwards on the right.
Looking along \(t_c\), opposite to its direction, \(t_c\) and \(x\) swapped position on the right diagram, which means given an anti-clockwise circulation, \(t_g\) lags \(x\). For the diagram on the left, \(t_g\) leads \(x\), if \(t_c\) is to point inwards, when the the particle is rotating anti-clockwise.
Do we have a problem here? Yes. In the post "Whacko and the Free Photons" dated 30 May 2014, EMW obeys Lenz's Law, photons fly off by themselves.
Given the direction of \(t_c\), the diagram on the left is a photon set free from its source, and the diagram on the right represents an EMW that collapses when power to the source is switched off.
We have a problem with negative time; \(t_c\,\, vs\,\,-t_c\).
It is possible to rotate in an anti-clockwise sense, but have \(t_c\) pointing in the opposite direction by moving in the \(-x\) direction for both cases. But the direction of \(t_c\) has real physical implications. \(t_c\) inwards is a EMW and \(t_c\) outwards is a photon. We can travel in the \(-x\) direction by reversing time; \(-t_T\).
This suggests that reversing one of \(t_c\), \(t_g\), or \(t_T\) reverses all time simultaneously. EMW then, is, relatively, in negative time. And so, has NEGATIVE energy!
All these notion of negative time and negative energy are avoided by instead adapting spins in clockwise or anti-clockwise sense; we can travel in the \(-x\) direction by simply reversing the sense of the circular motion; clockwise spin vs anti-clockwise spin.
Irrespective of which representation is valid, in all cases, EMWs and photons are, relatively, in opposite time of each other, positive energy vs negative energy, clockwise vs anti-clockwise spin and one being folded out into the other.
Can we choose which to produce, EMW or photons. Ooops, but first thing first, given any conservative field,
\(v^2+v^2_{tx}=c^2\)
where \(v\) is velocity in space and \(v_{tx}\) is velocity in the \(t_x\) time dimension. We can, by increasing the field, accelerate the particle such that,
\(v\to c\)
\(v_{tx}\to0\)
We can theoretically, by accelerating a basic particle to light speed with a spin, ie
to create a wave in space. Is this a Heat Wave?
It seems that EMW being different from photons complicates matters, slightly. We don't get photons/EMW in a straight forward manner. Given the left diagram and the post "Whacko and the Free Photons" dated 30 May 2014, an anti-clockwise spin produces a photon and an clockwise spin produces an EMW.
